Most rectifier front-end voltage source inverters do not allow power to flow back into the input AC supply. Such motor drives are said to operate either in the first quadrant or in the third quadrant as depicted in FIG. 1. However, in many applications, it is required to convert mechanical energy to electrical energy and ultimately channel it back to the AC supply. Such motor drives are said to be able to operate in all the four quadrants. When the duration of regeneration is short and the energy that needs to be regenerated is modest, dynamic braking units can be employed. In such cases, the motor drive operates in the second quadrant where the load is still rotating in the same direction but is being decelerated by the negative torque produced by the commanded frequency, which results in negative slip operation for induction motors (IM) or negative torque operation in permanent magnet (PM) motors. The mechanical energy is converted into electrical energy but this energy is dissipated as heat and hence such an operating mode is called a braking mode of operation.
Diode front-end drives operate only in the 1st and 3rd quadrants with marginal tolerance to operate in the second quadrant. On the other hand, active PWM converter front-end drives 10, see FIG. 2, also known as “line regenerative” drives can operate in all four quadrants. The most popular areas of use for active PWM converter drives is in wind power generation systems, large solar farms connected to the AC grid, elevators and escalators, pump jacks used in oil fields, and centrifuges. Here, the mechanical energy stored in the rotating load or phot energy from solar cells is converted into electrical energy and fed back into the AC grid.
Four-quadrant operation requires actively controlling the flow of power into and out of the AC supply 18 by channeling the current appropriately into the corresponding phases. For regeneration operation, the available electrical power is in the form of a DC voltage while the load is an AC power supply. The switches 20 in the PWM converter 12 perform the function of an inverter similar to the PWM inverter 14 at the output stage of the motor-drive unit. In other words, the voltage waveform at the AC source input terminal of the PWM converter 12 is a sinusoidal PWM voltage waveform with a sinusoidal fundamental component. However, the similarity ends there. In case of a typical motor-drive output stage, the load is an induction motor 16 and the voltage waveform is not critical. The motor 16 behaves like a large smoothing inductor and responds only to the fundamental component in the output voltage waveform. On the other hand, in case of the PWM converter 12, the load, looking out from the PWM converter 12 in to the supply, is not an induction motor. It is the sinusoidal AC source 18, typically of low impedance. Hence, the PWM waveform cannot be directly impressed across the sinusoidal AC supply, because large amplitudes of ripple currents would then flow into the AC source 18, which can potentially damage the semiconductor switches 20 in the PWM converter 12. For this reason, input LCL filters 22 are used to filter the switching frequency ripple. The main inductor that filters the PWM ripple current can be likened to the leakage inductances of electric motors. The remaining filter components are provided to ensure that the input sinusoidal source voltage remains ripple free.
During motoring, power flows from the input AC side to the output load side. If standard diode rectifiers were to be employed then the input diodes would not conduct until the instantaneous value of the input supply voltage went higher than the voltage across the DC bus capacitor 24. The conduction would cease soon after the input supply voltage went below the DC bus capacitor voltage. This would result in a pulsating current waveform, which is familiar to most. Since the PWM converter 12 is equipped with active switches 20, it is beneficial to force current to flow into and out of the AC source 18 even when the input supply voltage is lower than the DC bus voltage across the capacitor 24. This would result in continuous current flow and hence lower total current harmonic distortion. The concept can be understood if one disregards the presence of an AC source and assumes a pure inverter action forcing current to flow into a motor load with significant leakage inductance. The only difference is that in a typical inverter, the output can be of any desired frequency within the limits of the inverter, the motor, and the associated mechanical system, while in this case, the output frequency is fixed at the supply frequency of either 50 Hz or 60 Hz. The leakage inductance of the motor is replaced by an external inductor LD as explained earlier.
In a PWM converter 12, the input current has a different phase relationship with the input supply voltage depending on the quadrant of operation. In the first and third quadrants, the input line current is in phase with the line-neutral voltage, indicating positive power flow from the AC source 18 into the motor load 16 via the PWM converter 12. In the second and fourth quadrants, the line current is 180° phase-shifted with respect to the line-neutral voltage, indicating negative power flow or power flow from the load side into the AC source 18 via the PWM converter 12.
An added advantage of using a PWM converter front-end system is that during its operation, current can be made to flow into or out of the AC supply system at appropriate times in a continuous manner. This feature eliminates pulsating current that is typical of diode based AC to DC rectifiers feeding a large DC bus capacitor. By using an input LCL filter 22, the ripple in the input current is minimized resulting in almost sinusoidal current operation. The ability to shape the input current plays an important and significant role in channeling the mechanical energy, either from overhauling loads, dynamometers, or wind turbines, safely and reliably into the AC grid. The current and voltage waveforms can be made to comply with IEEE 519-1992 regulations by choosing an appropriately sized input LCL filter 22.
Whenever a capacitor based filter is introduced into a power system, there is bound to be interaction between the external harmonic source and the capacitive filter. Since in the LCL 22 filter being used here, the capacitor CY is an important and integral part of the filter 22, it often interacts with pre-existing harmonics in the power system. The pre-existing harmonic source could be due to other VFDs on the system or other PWM rectifiers connected to other loads on the same power source. In order to study the interaction of external harmonic source on the LCL filter 22, a single-phase equivalent circuit of the LCL filter 22 is developed. The single-phase equivalent circuit is derived from the viewpoint of the external harmonic source. The sequence of derivation is shown in FIG. 3.
In FIG. 3, all voltage sources are shorted. The entire LCL filter 22 is assumed to be powered from one single-phase harmonic source, VH. The harmonic source VH is assumed to be applied across line and neutral. DC voltage source and the other phase voltages are shorted to derive the single-phase equivalent circuit. The systematic progression of the single-phase equivalent circuit is shown in FIG. 3.
From FIG. 3, the expression for Z1 is derived next. The equivalent circuit is analyzed to compute the resonant conditions based on the filter component values.
                                                                        Z                1                            =                                                                    jω                    h                                    ·                                      L                    G                                                  +                                                      2                    ·                                          jω                      h                                        ·                                          L                      D                                        ·                                          (                                              2                        ⁢                                                  /                                                ⁢                                                  (                                                                                    jω                              h                                                        ·                                                          C                              Y                                                                                )                                                                    )                                                                                                  2                      ·                                              jω                        h                                            ·                                              L                        D                                                              +                                          (                                              2                        ⁢                                                  /                                                ⁢                                                  (                                                                                    jω                              h                                                        ·                                                          C                              Y                                                                                )                                                                    )                                                                                                                                                              Z                1                            =                                                                    jω                    h                                    ·                                      L                    G                                                  +                                                      2                    ⁣                                          ·                                              jω                        h                                            ·                                              L                        D                                                                                                  1                    -                                                                  ω                        h                        2                                            ·                                              L                        D                                            ·                                              C                        Y                                                                                                                                                                                    Z                1                            =                                                jω                  h                                ·                                  (                                                                                                              L                          G                                                ·                                                  (                                                      1                            -                                                                                          ω                                h                                2                                                            ·                                                              L                                D                                                            ·                                                              C                                Y                                                                                                              )                                                                    +                                              2                        ·                                                  L                          D                                                                                                            1                      -                                                                        ω                          h                          2                                                ·                                                  L                          D                                                ·                                                  C                          Y                                                                                                      )                                                                                                                                          Z                  1                                2                            =                                                jω                  h                                ·                                  (                                                                                                              L                          G                                                ·                                                  (                                                      1                            -                                                                                          ω                                h                                2                                                            ·                                                              L                                D                                                            ·                                                              C                                Y                                                                                                              )                                                                    +                                              2                        ·                                                  L                          D                                                                                                            2                      ·                                              (                                                  1                          -                                                                                    ω                              h                              2                                                        ·                                                          L                              D                                                        ·                                                          C                              Y                                                                                                      )                                                                              )                                                                                                                        Z                H                            =                                                jω                  h                                ·                                  (                                                                                    3                        ·                                                  L                          G                                                ·                                                  (                                                      1                            -                                                                                          ω                                h                                2                                                            ·                                                              L                                D                                                            ·                                                              C                                Y                                                                                                              )                                                                    +                                              2                        ·                                                  L                          D                                                                                                            2                      ·                                              (                                                  1                          -                                                                                    ω                              h                              2                                                        ·                                                          L                              D                                                        ·                                                          C                              Y                                                                                                      )                                                                              )                                                                                        (        3        )            
From equation (3), it is interesting to see that the value of the impedance ZH looking in from the AC source can go to zero when the numerator goes to zero. Equating the numerator in (3) to zero, yields the following results:
                                                        0              =                                                3                  ·                                      L                    G                                    ·                                      (                                          1                      -                                                                        ω                          h                          2                                                ·                                                  L                          D                                                ·                                                  C                          Y                                                                                      )                                                  +                                  2                  ·                                      L                    D                                                                                                                                          ω                h                2                            =                                                2                                      3                    ·                                          L                      G                                        ·                                          C                      Y                                                                      +                                  1                                                            L                      D                                        ·                                          C                      Y                                                                                                                                                              ω                h                2                            =                                                1                                                            L                      EFF                                        ·                                          C                      Y                                                                      ⁢                                                                  ⁢                where                ⁢                                                                  ⁢                                  L                  EFF                                ⁢                                                                  ⁢                is                ⁢                                                                  ⁢                defined                ⁢                                                                  ⁢                as                ⁢                                  :                                                                                                                        L                EFF                            =                                                3                  ·                                      L                    G                                    ·                                      L                    D                                                                                        2                    ·                                          L                      D                                                        +                                      3                    ·                                          L                      G                                                                                                                              (        4        )            
The possibility of resonance requires methods to damp it. Since the capacitor CY is involved in the resonance, adding a resistor RD (not shown) in series with the capacitor CY can help dampen oscillations. The value of the series resistor for an over-damped or critically damped system is derived next. The voltage equation for the equivalent loop consisting of LEFF, CY and RD is:
  v  =                    L        EFF            ·                        ⅆ          i                          ⅆ          t                      +          i      ·              R        D              +                  1                  C          Y                    ·              ∫                  i          ·                      ⅆ            t                              
Above equation has a natural response solution and a forced response solution.
Natural response part will yield the conditions for choice of damping resistor
                                                        0              =                                                                                          ⅆ                      2                                        ⁢                    i                                                        ⅆ                                          t                      2                                                                      +                                                                            R                      D                                                              L                      EFF                                                        ·                                                            ⅆ                      i                                                              ⅆ                      t                                                                      +                                  i                                                            L                      EFF                                        ·                                          C                      Y                                                                                                                                              0              =                                                (                                                            D                      2                                        +                                                                                            R                          D                                                                          L                          EFF                                                                    ·                      D                                        +                                          1                                                                        L                          EFF                                                ·                                                  C                          Y                                                                                                      )                                ·                i                                                                                                        D                1                            ,                                                D                  2                                =                                                                            -                                              R                        D                                                                                    2                      ·                                              L                        EFF                                                                              ±                                                            (                                                                                                    R                            D                            2                                                                                4                            ·                                                          L                              EFF                              2                                                                                                      -                                                  1                                                                                    L                              EFF                                                        ·                                                          C                              Y                                                                                                                          )                                                                                                                                                              R                D                2                            ≥                                                                    4                    ·                                          L                      EFF                                                                            C                    Y                                                  ⁢                                                                  ⁢                yields                ⁢                                                                  ⁢                an                ⁢                                                                  ⁢                over                ⁢                                  -                                ⁢                damped                ⁢                                                                  ⁢                or                ⁢                                                                  ⁢                critically                ⁢                                                                  ⁢                damped                ⁢                                                                  ⁢                                  system                  .                                                                                        (        5        )            
From (5), the value of the damping resistor RD in series with the filter capacitor CY should be greater than or equal to the critically damped resistance value given by: 2*√(LEFF/CY).
It should be noted that adding a resistor in series with the filter capacitor CY will increase the power loss in the filter, which can reduce the overall system efficiency. This may be a small price to pay to ensure that an external harmonic disturbance does not cause nuisance trips in the PWM converter 12. In this described topology (a resistor in series with the resonating capacitor), the effectiveness of the circuit is reduced since the series resistor adds impedance to the ripple current, which diminishes the capability of the circuit to shunt the ripple current. Further, the value of the series resistor should be large enough to provide sufficient damping but at the same time should be small to absorb the ripple current. This contradictory requirement makes the choice of a suitable capacitor-resistor combination difficult. Compromise may need to be made while selecting the value of the damping resistor either to improve damping or to reduce power loss in the damping resistors.
From the LCL filter topology shown in FIG. 2, it can be seen that a current path exists between the inductor LG and the capacitor CY even when the PWM converter switches 20 are OFF. This is a major source of power loss in the filter inductor LG and a damping resistor RD, if one is included. In addition, the insertion of a low pass filter in the form of LG-CY can create unwanted resonance conditions on the power grid, depending on other nonlinear loads preexisting on the grid. In view of the above situation, one can say that the standby power loss can become an important part of the overall grid power wastage when numerous PWM rectifiers in the form of wind power generating units or overhauling loads like in oil beam pumps form a major part of the power system.
This application is directed to improvements in use of LC filters to provide energy savings.